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Lines in the Space
Intersection of two lines

How to find the relationship between two lines in the space.

To find the relationship between the line

 A1x+B1y+C1z+D1=0 A2x+B2y+C2z+D2=0

and the line

 A3x+B3y+C3z+D3=0 A4x+B4y+C4z+D4=0

Form a system with the equations and calculate the ranks.

$\left{\fs1A_1x+B_1y+C_1z+D_1=0\\A_2x+B_2y+C_2z+D_2=0\\A_3x+B_3y+C_3z+D_3=0\\A_4x+B_4y+C_4z+D_4=0$

r = rank of the coefficient matrix
r'= rank of the augmented matrix

The relationship between the line and the plane can be described as follow:

 Case 1. Skew lines r=3 and r'=4 Case 2. Intersecting lines r=3 and r'=3 Case 3. Parallel lines r=2 and r'=3 Case 4. Coincident lines r=2 and r'=2

State the relationship between the following lines:

$r\;\equiv\;\frac{x-1}{2}=\frac{y}{1}=\frac{z-1}{1}\;$ and $s\;\equiv\;\frac{x}{1}=\frac{y}{-1}=\frac{z}{-1}\;$

Solution:

Form the system of equations and calculate the ranks.

$\left{x-2y-1=0\\y-z+1=0\\x+y=0\\x+z=0$

$M_1=\left(\begin{matrix}1&-2&0\\0&1&-1\\1&1&0\\\\1&0&1\end{matrix}\right)$
 1 -2 0 0 1 -1 1 1 0
$\neq\;0$ r=3

$M_2=\left(\begin{matrix}1&-2&0&1\\0&1&-1&-1\\1&1&0&0\\\\1&0&1&0\end{matrix}\right)$
 1 -2 0 1 0 1 -1 -1 1 1 0 0 1 0 1 0
$\neq\;0$ r'=4

They are skew lines.

State the relationship between the following lines:

 x+y+z-3=0 2x-y+z-2=0

and $s\;\equiv\;\frac{x-1}{2}=\frac{y-1}{1}=\frac{z-1}{3}\;$

Solution:

Form the system of equations and calculate the ranks.

$\left{x+y+z-3=0\\2x-y+z-2=0\\x-2y+1=0\\3x-2z-1=0$

$M_1=\left(\begin{matrix}1&1&1\\2&-1&1\\1&-2&0\\\\3&0&-2\end{matrix}\right)$
 1 1 1 2 -1 1 1 -2 0
$\neq\;0$ r=3

$M_2=\left(\begin{matrix}1&1&1&3\\2&-1&1&2\\1&-2&0&-1\\\\3&0&-2&1\end{matrix}\right)$
 1 1 1 3 2 -1 1 2 1 -2 0 -1 3 0 -2 1
=0 r=3

They are Intersecting Lines.

State the relationship between the lines:

 -5x+9y+2z=1 9x-5y+9z=6 2x-8y-4z=2 9x+9y-6z=-4

Solution:

1. Intersecting Lines
2. Skew lines
3. Coincident lines
4. Parallel lines